CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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$x \mapsto \sqrt{x}$ is a uniformly continuous bijection of $[0,\infty)$ onto itself whose inverse $x \mapsto x^2$ is not uniformly continuous

Statement refuted

Refuted claim: the inverse of a uniformly continuous bijection of metric spaces is uniformly continuous (Uniform continuity of a map of metric spaces: one $\delta$ serving every point, Injection, surjection, bijection).

Let $X := [0,\infty) \subseteq \mathbb{R}$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length) with the metric inherited from the real line (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset), and let

$g : X \to X, \quad g(x) := \sqrt{x}, \qquad h : X \to X, \quad h(x) := x^{2}.$

Then $g$ is a bijection of $X$ onto itself with inverse $h$ ; $g$ is uniformly continuous; and $h$ is not uniformly continuous, the pairs $n + 1/n$ and $n$ defeating every candidate $\delta$ at $\varepsilon = 2$ .

Facts & Assumptions

Given: $X = [0,\infty)$ with the metric inherited from $\mathbb{R}$ ; the maps $g(x) = \sqrt x$ and $h(x) = x^2$ ; a natural $n \ge 1$ ; reals $\varepsilon, \delta > 0$ .

[L1]

Every $a \ge 0$ has a unique $\sqrt a \ge 0$ with $(\sqrt a)^2 = a$ (Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$ ; the positives are $\{x^2 : x \neq 0\}$ , Integer powers $a^m$ ).

[L2]

For $a,b \ge 0$ : $a \le b$ if and only if $a^2 \le b^2$ (Squaring is monotone on the nonnegatives).

[L3]

$\sqrt{\cdot}$ on $[0,\infty)$ is $1/2$ -Hölder with constant $1$ and is uniformly continuous ( $\sqrt{\cdot}$ on $[0,\infty)$ is uniformly continuous and exactly $1/2$ -Hölder, and is not Lipschitz, Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction, Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent).

[L4]

Factorisation of a difference of squares: $x^2 - y^2 = (x-y)(x+y)$ (Factorisation of $b^n - a^n$ , and the resulting Lipschitz estimate).

[L5]

For every real $\eta > 0$ there is a natural $n \ge 1$ with $1/n < \eta$ ; positive naturals are positive reals; and $a > 0$ gives $1/a > 0$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Every complete ordered field is Archimedean, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L6]

The absolute value and the usual metric of $\mathbb{R}$ (Basic properties of the absolute value, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded, Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L7]

Uniform continuity: one $\delta > 0$ per $\varepsilon > 0$ serving every pair (Uniform continuity of a map of metric spaces: one $\delta$ serving every point).

Counterexample

technique · direct

1.1

$g$ and $h$ map $X$ into $X$ : $\sqrt x \ge 0$ for $x \ge 0$ by [L1], and $x^2 \ge 0$ for every real $x$ .

L1L2

1.2

$g$ is uniformly continuous on $X$ .

L3L7

1.3

For every real $t > 0$ and every $x,y \ge 0$ : $h(x) - h(y) = x^2 - y^2 = (x-y)(x+y)$ .

2.1

$h(g(x)) = (\sqrt x)^2 = x$ for every $x \ge 0$ , and $g(h(x)) = \sqrt{x^2} = x$ for every $x \ge 0$ , the latter because $x \ge 0$ and $x^2 = x^2$ make $x$ the unique nonnegative square root of $x^2$ . So $g$ and $h$ are mutually inverse bijections of $X$ onto itself.

step 1.1L1L2

2.2

Fix a real $\delta > 0$ and take a natural $n \ge 1$ with $1/n < \delta$ ; put $x := n + 1/n$ and $y := n$ , both in $X$ . Then $|x - y| = 1/n < \delta$ .

step 1.3L5L6

3.1

But $h(x) - h(y) = (x-y)(x+y) = \dfrac{1}{n}\Big(2n + \dfrac1n\Big) = 2 + \dfrac{1}{n^2} > 2$ . So the pair $(x,y)$ satisfies $|x-y| < \delta$ and $|h(x)-h(y)| > 2$ .

step 1.3step 2.2L5L6

4.1

Since $\delta > 0$ was arbitrary, no $\delta$ witnesses the uniform continuity condition for $h$ at $\varepsilon = 2$ , so $h$ is not uniformly continuous.

step 2.2step 3.1L7

5.1

Therefore $g$ is a uniformly continuous bijection of $X$ onto itself whose inverse $h$ is not uniformly continuous, which refutes the claim above.

step 2.1step 1.2step 4.1∎

Remarks

Both maps are continuous, and one direction of the pair is even Hölder. What fails is only the uniformity of the inverse: $h$ stretches distances by the factor $x + y$ , which is unbounded on $[0,\infty)$ , so no single $\delta$ can serve every pair. On any bounded piece $[0,M]$ the same factor is at most $2M$ , so $h$ is Lipschitz there and the phenomenon disappears (Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction).
The witnesses shrink and their images do not. The pairs $n + 1/n$ and $n$ are at distance $1/n$ , which tends to $0$ , while their images stay more than $2$ apart. That is the shape of every failure of uniform continuity: a family of pairs whose separation vanishes and whose image separation does not.
Indexing. The witnesses are indexed by naturals $n \ge 1$ , because $1/n$ appears; $\mathbb{N}$ contains $0$ in this library and $1/0$ does not exist.
A homeomorphism can be uniformly continuous in one direction only. So "uniformly homeomorphic" is a genuinely stronger relation than "homeomorphic", and this pair shows the two differ; the metric-level version of the same point is the gap between topological and uniform equivalence (Topologically, uniformly and Lipschitz equivalent metrics on a set, On $(0,\infty)$ the metrics $|x-y|$ and $|1/x - 1/y|$ share their topology and not their Cauchy sequences).

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 100 results over 31 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Uniform continuity (Wikipedia) (standard reference, not scraped)
Nth root (Wikipedia) (standard reference, not scraped)

x↦xx \mapsto \sqrt{x}x↦x​ is a uniformly continuous bijection of [0,∞)[0,\infty)[0,∞) onto itself whose inverse x↦x2x \mapsto x^2x↦x2 is not uniformly continuous